Is this true :${(a+ib)}^{(k+ij)}=0$ iff $0<a=k<1$ and $b<j$?

Question

let $z=a+ib ,s=k+ij$ are two complex numbers and let $f(z,s)$ be a complex

function defined as follow :$$f(z,s)=z^s={(a+ib)}^{(k+ij)}$$ and $a,b,j, k$ are non -nul real numbers .

.After some calculations in wolfram alpha which i performed I got this:

$f(z,s)$ approach to zero iff $0<a=k<1$ and $b<j$ .

My question here: Is "$f(z,s)=0$ iff $0<a=k<1$ and $b<j$"really a true claim ?

Thank you for any help